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DENIUS [597]
3 years ago
11

A certain factory whistle can be heard up to a distance of 2.5 km. Assuming that the acoustic output of the whistle is uniform i

n all directions, how much acoustic power is emitted by the whistle? The threshold of human hearing is 1.0 × 10-12 W/m2.
Physics
1 answer:
enyata [817]3 years ago
4 0

Answer:

Emitted power will be equal to 7.85\times 10^{-5}watt

Explanation:

It is given factory whistle can be heard up to a distance of R=2.5 km = 2500 m

Threshold of human hearing I=10^{-12}W/m^2

We have to find the emitted power

Emitted power is equal to P=I\times A

P=I\times 4\pi R^2

P=10^{-12}\times 4\times 3.14\times  2500^2=7.85\times 10^{-5}watt

So emitted power will be equal to 7.85\times 10^{-5}watt

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Bas_tet [7]

Given info

d = 0.000250 meters = distance between slits

L = 302 cm = 0.302 meters = distance from slits to screen

\theta_8 = 1.12^{\circ} = angle to 8th max (note how m = 8 since we're comparing this to the form \theta_m)

x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters} (n = 5 as we're dealing with the 5th minimum )

---------------

Method 1

d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}

Make sure your calculator is in degree mode.

-----------------

Method 2

\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\

-----------------

Method 3

\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\

There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.

7 0
3 years ago
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